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19 Sep 2007, 09:50

b14kumar wrote:

b14kumar wrote:

Fistail wrote:

GK_Gmat wrote:

Is A positive?

1) x^2 - 2x +A is positive for all x2) A*x^2 + 1 is positive for all x

Pls. explain. Thanks.

1: x^2 - 2x +A > 0

refrese the inequality as : x^2 - 2x + 1 + A -1 > 0so it is reduced to (x - 1)^2 + A – 1 > 0.now (x -1)^2 can be 0 or grater than 0. if it is 0, A - 1 has to be +ve and to be so, A has to be grater than 1. so suff.

2: A*x^2 + 1 is positive for all x.

it is clearly insufficient because x^2 is 0, A could be +ve or -ve and the expression still is +ve..

so A works here.

Indeed a good approach Fistail.

But how about putting the value of x = -1 and obtaining the possible values of A as I mentioned in my above earlier post?

Writing again:

ST1: x^2 - 2x +A is positive for all x

Let's say x = -1

Then, 1 + 2 + A > 0=> 3 + A > 0=> A > (-3)

So A may be -2 , -1 , 0 or > 0

Hence A may be positive or negative.

- Brajesh

I again looked at your approach.

As per you:

refrese the inequality as : x^2 - 2x + 1 + A -1 > 0so it is reduced to (x - 1)^2 + A – 1 > 0.now (x -1)^2 can be 0 or grater than 0. if it is 0, A - 1 has to be +ve and to be so, A has to be grater than 1. so suff.Why are you assuming only the case when (x-1)^2 is equal to 0?

Well, (x -1)^2 will always be >= 0 but it does not mean that "A – 1" has to be positive for all the cases.Imagine, (x -1)^2 is equal to 4 (by taking x = 3 ) , in this case, (A-1) can be (-3) i.e A can be (-2) and still the whole inequality will be intact.

Please let me know your opinion.

- Brajesh

go for the extream case. make (x^2 - 1) minimum. for other values of (x - 1)^2, its not a problem but when (x - 1)^2 is lowest what happens to the value of A is important? since A must be +ve when (x - 1)^2 is minimum, the A is +ve.